If $ \limsup ( a_n ) = L \in \mathbb{R} $, then is $ (a_n) $ bounded from above?

Question

If $ \limsup ( a_n ) = L \in \mathbb{R} $ ( Meaning the largest partial limit of $ (a_n) $ is a finite number ) then does this mean $ ( a_n) $ is bounded from above?

I had this question in my mind for awhile and I think the answer is no but I couldn't think of a counter-example.

If it's not bounded from above, isn't there a subsequence that goes to $\infty$? — saulspatz, Apr 24 '21 at 16:05
How can there be such subsequence that goes to $ \infty $ if the largest partial limit is a $ L \in \mathbb{R} $? — hazelnut_116, Apr 24 '21 at 16:10
$\exists N$ such that $ a_n \le L+1$ for $n \ge N$. Can you derive an upper bound for all $a_n$? — Martin R, Apr 24 '21 at 16:24
@MartinR , Yes the upper bound is: $ M = max{a_{1},a_{2},...,a_{N},L+1} $. The proof by contradiction introduced below in the answers cleared things for me. — hazelnut_116, Apr 24 '21 at 16:43

score 3 · Accepted Answer · answered Apr 24 '21 at 16:04

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If $(a_{n})$ is not bounded from above, then there is a subsequence $(a_{n_{k}})$ such that $a_{n_{k}}\rightarrow\infty$, but then $\limsup a_{n}\geq\lim_{k}a_{n_{k}}=\infty$.

answered Apr 24 '21 at 16:04

user284331

55,591

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So we get a contradiction, hence $ ( a_n) $ must be bounded above? – hazelnut_116 Apr 24 '21 at 16:11
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Yes, that is right. – user284331 Apr 24 '21 at 16:11

If $ \limsup ( a_n ) = L \in \mathbb{R} $, then is $ (a_n) $ bounded from above?

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